Let $\angle PAB=\alpha$, $\angle DPA=\beta$ and $PQ$ be a common tangent to $\omega_1$ and $\omega_2$, where $Q$ lies on $AB$. Since $AC$ and $PQ$ is tangent to $\omega_1$, then $\angle PAB=\angle ADP=\angle APQ=\alpha$ and $\angle DAC=180-\beta$. $\angle DCA=180-\angle DAC-\angle ADP=\beta-\alpha$. Since $PQ$ is tangent to $\omega_2$, then $\angle DCA=\angle BPQ=\beta-\alpha$. $\angle APB=\angle APQ+\angle BPQ=DPA=\alpha+\beta-\alpha=\beta$.

$AC$ is tangent to $(APD) \implies \triangle ACD \sim \triangle PCD \implies \angle APD=180-\angle DAC=\angle ADC +\angle ACD$. Sine $(APD)$ and $(BPC)$ are externally tangent $\angle APB=\angle ADC+\angle ACD=\angle APD$.

Chase The Angle In the diagram, probably you can figure out the solution. Try proving that the same coloured angles are same ($PQ$ is the common internal tangent of $\omega_1$ and $\omega_2$. If you can show that the yellow marked angles are the difference of brown and green marked angles, then the proof follows.

Attachments:

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -44.967368876545386, xmax = 57.74582857537591, ymin = -43.370296696745584, ymax = 20.808668508506738; /* image dimensions */ /* draw figures */ draw(circle((-9.752639984581936,-9.043405498003121), 9.630659266346862), linewidth(0.8) + red); draw(circle((19.117650960792165,-9.430061180307236), 19.242220775090086), linewidth(0.8) + red); draw((-12.543800780931146,0.17391620156861087)--(19.28248541871449,9.811453571883922), linewidth(0.8) + blue); draw((19.28248541871449,9.811453571883922)--(-9.835139012015297,-18.673711402539347), linewidth(0.8) + blue); draw((-9.835139012015297,-18.673711402539347)--(-12.543800780931146,0.17391620156861087), linewidth(0.8) + blue); draw((-12.543800780931146,0.17391620156861087)--(24.694437635179472,-27.846426477120957), linewidth(0.8) + blue); draw((24.694437635179472,-27.846426477120957)--(19.28248541871449,9.811453571883922), linewidth(0.8) + blue); draw((8.306025046532069,6.487595853546962)--(-0.12284432110230803,-9.172375975639008), linewidth(0.8) + blue); /* dots and labels */ dot((-0.12284432110230803,-9.172375975639008),dotstyle); label("$P$", (0.1458786317494579,-8.528368933643124), NE * labelscalefactor); dot((-12.543800780931146,0.17391620156861087),dotstyle); label("$A$", (-12.273690994790046,0.8702242972515678), NE * labelscalefactor); dot((8.306025046532069,6.487595853546962),linewidth(4pt) + dotstyle); label("$B$", (8.604612539554742,7.046442706125222), NE * labelscalefactor); dot((19.28248541871449,9.811453571883922),linewidth(4pt) + dotstyle); label("$C$", (19.547260372667925,10.335950336938364), NE * labelscalefactor); dot((-9.835139012015297,-18.673711402539347),linewidth(4pt) + dotstyle); label("$D$", (-9.588378643105829,-18.12836059091413), NE * labelscalefactor); dot((24.694437635179472,-27.846426477120957),linewidth(4pt) + dotstyle); label("$A'$", (24.985017884828466,-27.325555395432506), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Quick exercise for spiral similarity $\color{black}\rule{25cm}{1pt}$ We want to show that $APD \sim BPA$, since we have that $\angle PDA = \angle BAP$, we just want to show that $\angle DAP = \angle ABP$. Let $A'$ be the point of the second intersection of $AP$ with $\omega_2$. The we have a spiral similarity $\varphi$ centered at $P$. This implies that $\angle PAD = \angle PA'C = \angle PBA$, thus we are done

Let $E$ be the second intersection of $AP$ with $\omega_2.$ By homothety at $P$ sending $\omega_1$ to $\omega_2$ we have $AD||CE,$ so \[\angle DAP=\angle CEP= \angle ABP,\angle PAB=\angle PDA\implies\triangle PBA\stackrel{+}{\sim}\triangle PAD\implies \angle BPA=\angle APD\]as desired. $\blacksquare$ wait i just realized this is identical to above lol